25.0 mL of 0.050 M Ba(NO₃)₂ is mixed with 25.0 mL of 0.020 M NaF . K_(Sp) of BaF₂ is 0.5 × 10⁻⁶ at 298 K. The ratio of [Ba²⁺][F⁻]² and Kₛₚ is ___________.
(Nearest integer)

Solution

Chemistry ma’am · 4 steps

Mixing doubles the volume, so halve both concentrations before you build the ionic product.

1. Dilute on mixing. Equal volumes of 25.0 mL give a total of 50.0 mL, so each concentration is halved. [Ba²⁺] = 0.050/2 = 0.025 M and [F⁻] = 0.020/2 = 0.010 M.
2. Ionic product. For BaF₂ ⇌ Ba²⁺ + 2F⁻, the ionic product uses the square of the fluoride concentration.

   [Ba²⁺][F⁻]² = 0.025 × (0.010)² = 2.5 × 10⁻⁶

3. Compare with Ksp. (2.5 × 10⁻⁶)/(0.5 × 10⁻⁶) = 5. The ionic product is larger than Kₛₚ, so BaF₂ will precipitate.
4. So. The required ratio is 5.

The common slip: 40. Using the original concentrations without dilution gives 0.050 × (0.020)² = 2 × 10⁻⁵ and a ratio of 40, not 5.

The idea to keep. Whenever two solutions are mixed, first find each ion's new concentration in the total volume. Then compare the ionic product with Kₛₚ, raising each ion to its stoichiometric power.

Answer from the NTA final answer key. Solution by PrepWiser.